00:01
Use newtuce method to find all solution of the equation correct to the 6 decimal point.
00:11
The question is square root of x plus 1 is equal to x square minus x.
00:23
So it can be written as x square minus x plus 1 square root is equal to 0.
00:33
So here let f x is equal to the complete equation we got above that is x square minus x minus x plus 1 is equal to 0.
00:46
Now we will put f minus 1 is equal to 1 plus 1 minus 0 that is 2.
00:55
We will find f of 0 which is minus 1 and f of 1 is equal to minus root 2 simply by putting the values.
01:06
Fine and if you put f let's find f of two then f of two will get two minus root three so one of the root lies in one of the root lies in minus one comma zero and the second one second root lies in interval one comma two so x 1 plus 1th is equal to x n minus f of xn minus f dash of x n.
02:00
This is by definition is equal to x n is equal to x square minus x n, x n upon derivative of this complete equation.
02:15
That is 2 x n minus 1 minus 1 upon 2 root x n plus 1 so it can be written as 2 x square x n par under root xn plus 1 plus xn plus 2 upon 4 xn minus 2 multiplied by x n plus 1 plus 1 minus 1 and let's get it a name as g of x n now consider x not consider x not is equal to 0 .5 minus 0 .5 as the initial approximation for the root in minus 1 .0 so x of 1 will get g of x not is equal to g of minus 0 .0 is equal to g of minus 0.
03:26
5 is equal to minus 4 .4 -8415.
03:33
Now x2 if you calculate that is g of x1 is equal to minus 0 .484028.
03:47
Fine...